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Question 794405: Find three POSITIVE numbers a, b and c such that
1 times a plus 7 times b plus 4 times c equals 177,
and such that the product of a, b and c is a MAXIMUM.
Answer by AnlytcPhil(1807) (Show Source):
You can put this solution on YOUR website!
Let a, b, and c be the three dimensions of a rectangular
solid with dimensions a by b by c.
The product of a, b and c will represent the volume of
that rectangular solid. We know that the volume
of a rectangular solid is maximized when the
rectangular solid is a cube. That would be when
a=b=c. If such a cube were possible, then that would
give us the maximum product abc.
Let's see if they could all be equal, i.e., a = b = c.
If a = b = c then
1a + 7b + 4c = 177 gives
1a + 7a + 4a = 177
12a = 177
a = 14.75
Of course a, b, and c must be integers, and cannot
be 14.75, but that was not done in vain, because the
rectangular solid that is nearest to a 14.75×14.75×14.75
cube will have the maximum volume abc.
So we want to take a,b, and c as close to 14.75
as possible.
The nearest integer to 14.75 is 15. So we will see
if it is possible to make two of the dimensions of the
rectangular solid 15 and the third dimension be an
integer.
We try a=b=15, but when we substitute those into 1a + 7b + 4c = 177
we get c = 57/4. So that is not possible.
We try a=c=15, but when we substitute those into 1a + 7b + 4c = 177
we get b = 102/7. So that is also not possible.
We try b=c=15, and when we substitute those into 1a + 7b + 4c = 177
we get c = 12. Aha! That IS the answer!
That is the nearest we can get to a 14.75×14.75×14.75 cube, so the
solution
a=12, b=15, and c=15, gives a maximum volume of a·b·c = 2700
Edwin
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